Question

grafica la linea: $y + 1 = -\frac{1}{3}(x - 2)$

Explanation:

Step1: Identify the form of the line

The equation given is \( y + 1 = -\frac{1}{3}(x - 2) \), which is in point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(2,- 1) \) and the slope \( m =-\frac{1}{3}\).

Step2: Find a second point

Using the slope \( m=-\frac{1}{3}\), from the point \( (2,-1) \), we can move 3 units to the right (increase \( x\) by 3) and 1 unit down (decrease \( y\) by 1). So \( x = 2+3 = 5\) and \( y=-1 - 1=-2 \). So another point on the line is \( (5,-2) \).

Step3: Plot the points and draw the line

We plot the points \( (2,-1) \) and \( (5,-2) \) (or other points found using the slope) on the coordinate plane and then draw a straight line passing through them. The given graph in the problem seems to have points plotted. Let's check the intercepts as well. To find the \( y\) - intercept, set \( x = 0\) in the equation \( y+1=-\frac{1}{3}(0 - 2)\), \( y + 1=\frac{2}{3}\), \( y=\frac{2}{3}-1=-\frac{1}{3}\). To find the \( x\) - intercept, set \( y = 0\) in the equation \( 0 + 1=-\frac{1}{3}(x - 2)\), \( 1=-\frac{1}{3}(x - 2)\), \( x - 2=-3\), \( x=-1\). So the line should pass through \( (-1,0) \), \( (2,-1) \), \( (5,-2) \) etc. The graph provided has points that seem to align with the line defined by the equation \( y+1 =-\frac{1}{3}(x - 2) \) (rewritten as \( y=-\frac{1}{3}x+\frac{2}{3}-1=-\frac{1}{3}x-\frac{1}{3}\)).

Answer:

To graph the line \( y + 1=-\frac{1}{3}(x - 2) \), we can use the point - slope form to identify a point \( (2,-1) \) and the slope \( -\frac{1}{3}\) to find other points. Then plot the points and draw a straight line through them. The given graph in the problem appears to be the correct graph of the line (with points plotted as per the equation's characteristics).